Mathematics for the interested outsider

Okay, let’s dive right in with a first step towards proving the inverse function theorem we talked about at the end of yesterday’s post. This is going to get messy.

We start with a function and first ask that it be continuous and injective on the closed ball of radius around the point . Then we ask that all the partial derivatives of exist within the open interior — note that this is weaker than our existence condition for the differential of — and that the Jacobian determinant on . Then I say that the image actually contains a neighborhood of . That is, the image doesn’t “flatten out” near .

The boundary of the ball is the sphere of radius :

Now the Heine-Borel theorem says that this sphere, being both closed and bounded, is a compact subset of . We’ll define a function on this sphere by

which must be continuous and strictly positive, since if then , but we assumed that is injective on . But we also know that the image of a continuous real-valued function on a compact, connected space must be a closed interval. That is, , and there exists some point on the sphere where this minimum is actually attained: .

Now we’re going to let be the ball of radius centered at . We will show that , and is thus a neighborhood of contained within . To this end, we’ll pick and show that .

So, given such a point , we define a new function on the closed ball by

This function is continuous on the compact ball , so it again has an absolute minimum. I say that it happens somewhere in the interior .

At the center of the ball, we have (since ), so the minimum must be even less. But on the boundary , we find

so the minimum can’t happen on the boundary. So this minimum of happens at some point in the open ball , and so does the minimum of the square of :

Now we can vary each component of separately, and use Fermat’s theorem to tell us that the derivative in terms of must be zero at the minimum value . That is, each of the partial derivatives of must be zero (we’ll come back to this more generally later):

This is the product of the vector by the matrix . And the determinant of this matrix is : the Jacobian determinant at , which we assumed to be nonzero way back at the beginning! Thus the matrix must be invertible, and the only possible solution to this system of equations is for , and so .

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.